lim-x-n-n-n-1-n-lim-x-0-tan-4x-tan-4x-

Question Number 157216 by Khalmohmmad last updated on 21/Oct/21

\lim_{x \to \infty} {(\frac{n!}{n^{n}})}^{\frac{1}{n}} = ?

\lim_{x \to 0} \frac{\sqrt{\tan 4 x}}{\tan \sqrt{4 x}} = ?

Commented by cortano last updated on 21/Oct/21

(2) \lim_{x \to 0^{+}} \frac{\sqrt{\tan 4 x}}{\tan \sqrt{4 x}} = \lim_{x \to 0^{+}} \frac{\frac{\sqrt{\tan 4 x}}{\sqrt{4 x}}}{\frac{\tan \sqrt{4 x}}{\sqrt{4 x}}}

= \lim_{x \to 0^{+}} \frac{\sqrt{\frac{\tan 4 x}{4 x}}}{\frac{\tan \sqrt{4 x}}{\sqrt{4 x}}} = 1

Answered by puissant last updated on 21/Oct/21

1)

L = \lim_{n \to \infty} {(\frac{n!}{n^{n}})}^{\frac{1}{n}}; l n (L) = \lim_{n \to \infty} \frac{1}{n} l n (\frac{n!}{n^{n}})

= \lim_{n \to \infty} \frac{1}{n} {l n (n!) - n l n (n)}

= \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} {l n (k) - l n (n)}

= \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} l n (\frac{k}{n}) = \int_{0}^{1} l n (x) d x

I B P \to {\begin{cases} u = l n x \\ v^{'} = 1 \end{cases} \Rightarrow {\begin{cases} u^{'} = \frac{1}{x} \\ v = x \end{cases}

\Rightarrow l n (L) = {[x l n x]}_{0}^{1} - \int_{0}^{1} 1 d x

\Rightarrow l n (L) = - 1 \to L = e^{- 1} = \frac{1}{e} . .

∴∵ L = \lim_{n \to \infty} {(\frac{n!}{n^{n}})}^{\frac{1}{n}} = \frac{1}{e} . .

\dots \dots \dots . . L e p u i s s a n t \dots \dots \dots \dots

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